Bob's average golf score at his local course is 93.6 from a random sample of 34 rounds of golf. Assume that the population standard deviation for his golf score is 4.2. The 90% confidence interval around this sample mean is ________.
A) [89.0, 98.3]
B) [90.5, 96.7]
C) [92.4, 94.8]
D) [93.2, 94.0]

Question

Quizplus · Accepted Answer

We use the formula for a confidence interval for a population mean:
CI = x̄ ± z*(σ/√n)
where x̄ is the sample mean, z is the z-score corresponding to the confidence level (1.645 for 90%), σ is the population standard deviation, and n is the sample size.

Plugging in the given values, we get:
CI = 93.6 ± 1.645*(4.2/√34)
CI = 93.6 ± 1.096
CI = [92.504, 94.696]

Rounding to one decimal place gives us the answer of [92.4, 94.8], which corresponds to choice C.

Bob's Average Golf Score at His Local Course Is 93